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I'm looking at an equation for average power in a signal

$$ p_{avg} = \frac{1}{R} v_{rms}^2 $$

and wondering why it isn't

$$ p_{avg} = \frac{1}{R} |v|_{avg}^2 $$

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    \$\begingroup\$ Because the square of the average is not always the average of the squares, not even for positive numbers. 0 and 10 average to 5, square that to get 25. But the average of their squares (0 and 100) is 50. Not even close! Why the square in the first place? Power is Voltage * current, but the current is itself proportional to the Voltage, so the power is proportional to the Voltage squared. \$\endgroup\$ – Wouter van Ooijen Sep 14 '12 at 20:14
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Simple: the average of a sine is zero.

Power is proportional to voltage squared:

\$ P = \dfrac{V^2}{R} \$

so to get average power you calculate average voltage squared. That's what the RMS refers to: Root Mean Square: take the square root of the average (mean) of the squared voltage. You have to take the square root to get the dimension of a voltage again, since you first squared it.

enter image description here

This graph shows the difference between the two. The purple curve is the sine squared, the yellowish line the absolute value. The RMS value is \$\sqrt{2}/2\$, or about 0.71, the average value is \$2/\pi\$, or about 0.64, a difference of 10 %.

RMS gives you the equivalent DC voltage for the same power. If you would measure the resistor's temperature as a measure of dissipated energy you'll see that it's the same as for a DC voltage of 0.71 V, not 0.64 V.

edit
Measuring average voltage is cheaper than measuring RMS voltage however, and that's what cheaper DMMs do. They presume the signal is a sine wave, measure the rectified average and multiply the result by 1.11 (0.71/0.64) to get the RMS value. But the factor 1.11 is only valid for sinewaves. For other signals the ratio will be different. That ratio got a name: it's called the signal's form factor. For a 10 % duty cycle PWM signal the form factor will be \$1/\sqrt{10}\$, or about 0.316. That's a lot less than the sine's 1.11. DMMs which are not "True RMS" will give large errors for non-sinusoidal waveforms.

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  • \$\begingroup\$ To your first point, I edited my second equation to use the average absolute value, which is what I meant. What I'm not seeing is why the order of the two operations (average and square) matters. Average voltage squared, vs average squared voltage. \$\endgroup\$ – Rob N Sep 14 '12 at 17:13
  • \$\begingroup\$ Because of the square-law relationship the average of the power and the average of the voltage are two very different things. \$\endgroup\$ – Dave Tweed Sep 14 '12 at 17:27
  • \$\begingroup\$ @RobN, the instantaneous power is \$p(t) = v^2(t)/R \$. The average power is the time average of \$p(t)\$. Thus, the average power is proportional to the time average of the squared voltage. Also, the order matters because the average of the squares is not equal to the square of the average. \$\endgroup\$ – Alfred Centauri Sep 14 '12 at 19:33
  • \$\begingroup\$ Note that the average of the square of a sine is one half. The inverted and phase shifted curve fits exactly into the valleys in the original curve, a consequence of Pythagoras' law, and their sum is a constant 1. \$\endgroup\$ – starblue Sep 14 '12 at 19:43
  • \$\begingroup\$ Sorry about offtopic guys, but how can I draw graphs like this with minimum effort? By graphs like this I mean some sin, |sin| etc. \$\endgroup\$ – Kamil Nov 29 '14 at 20:43
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Now speaking in terms of equations:

\$ P_{avg}= avg(P_{inst}) \$

Now \$P_{inst} = v(t) \cdot i(t)\$ where \$v(t)\$ and \$i(t)\$ are instantaneous voltage and current resp. Hence

\$ P_{inst} = \dfrac{(v(t))^2}{R} \$

\$ P_{avg} = avg(\dfrac{((v(t))^2}{R}) \$

\$ P_{avg}= \dfrac{V_{rms}^2}{R} \$

As RMS = \$\sqrt{\text{average of squares of inst.}}\$

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    \$\begingroup\$ So? All you have presented is equations, without explanation or argument. This is not useful. \$\endgroup\$ – Chris Stratton Nov 14 '18 at 4:56
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The why is simple.

You want 1 W = 1 W.

Imagine a primitive heater, a 1 ohm resistor.

Consider 1 VDC into a 1 ohm resistor. Power consumption is obviously 1 W. Do that for one hour, and you burn one watt-hour, generating heat.

Now, instead of DC, you want to feed AC to the resistor, and produce the same heat. What AC voltage do you use?

It turns out that RMS voltage gives you the result you want.

THAT'S why RMS is defined the way it is, to make the power numbers come out right.

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    \$\begingroup\$ This has hints of a useful answer, but it must be all but entirely rewritten to make them clear \$\endgroup\$ – Chris Stratton Nov 14 '18 at 4:57
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Because the power equal to V^2/R so that you calculate the average of the squared voltages along the sinusoidal wave to get V^2avg. For simplicity we take the average of this mean then we can deal with it as we wish.

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  • \$\begingroup\$ This is essentially the key point, but it could be explained in a much better way. \$\endgroup\$ – Chris Stratton Nov 14 '18 at 4:57
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The answer is the reason given by John R. Strohm and the explanation is as follows: (requires a few additions to stevenvh's answer)

You see when you send a DC through a resistor and an AC wave through a resistor the resistor get heated up in both cases, but according to the equation for the average value the heating effect for ac should be 0 but its not why? This is because the when the electrons move in a conductor they strike atoms and this energy imparted to the atoms are consequently felt as heat, now AC does the same thing only the the electrons are moving in different directions but the energy transfer here is independent of the direction and so the conductor heats up all the same.

When we find the average value the ac components are cancelled out and hence fail to explain why the heat is generated but RMS equation rectifies that - as stevenvh says by taking the square and then the square root we are transposing the negative portion to the top of the axis such that the positive and negative portions don't cancel off.

This is why that we say that the average and the RMS values of a DC wave is the same.

The same applies to any real world signal (by this I mean imperfect - not pure AC) as Fourier series says that any wave can be replaced by a correct combination of sine and cosine waves and since the frequencies of the waves are higher (integer multiples of the base frequency) they too get cancelled out, isolating the DC component.

The above is the reason that we define RMS value as the equivalent value of DC that generates the same amount of heat as the AC wave.

Hope this helps.

PS: I know that the explanation for how heat is generated is quite ambiguous but I am at loss to find a better one, I went with it anyway because it helps convey the message

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  • \$\begingroup\$ There are some useful points here, but this is far too chatty; to be a good answer you must entirely re-write this in a factual way. \$\endgroup\$ – Chris Stratton Nov 14 '18 at 4:58
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y(x) = |x| is not differentiable, because y'(0) is undefined.

y(x) = sqrt(x * x) is differentiable.

However they are otherwise equivalent.


Vrms = average(abs(v(t))) = average(sqrt(v(t) * v(t)))

Why did they pick one definition over the other? Well, one is an average of a differentiatable function.

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    \$\begingroup\$ That's not why, though. It's because using the RMS voltage gives you the same average power as if you calculated the instantaneous power at each point and then averaged it. This also holds for current. All of the equations for DC behavior hold exactly for AC, if and only if the RMS value is used. \$\endgroup\$ – Hearth May 27 at 0:40

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